## Barron’s SAT Subject Test Math Level 2, 10th Edition (2012)

MODEL TEST 5

### Model Test 5

 The following directions are for the print book only. Since this is an e-Book, record all answers and self-evaluations separately.

Tear out the preceding answer sheet. Decide which is the best choice by rounding your answer when appropriate. Blacken the corresponding space on the answer sheet. When finished, check your answers with those at the end of the test. For questions that you got wrong, note the sections containing the material that you must review. Also, if you do not fully understand how you arrived at some of the correct answers, you should review the appropriate sections. Finally, fill out the self-evaluation chart in order to pinpoint the topics that give you the most difficulty.

 50 questions: 1 hour Directions: Decide which answer choice is best. If the exact numerical value is not one of the answer choices, select the closest approximation. Fill in the oval on the answer sheet that corresponds to your choice. Notes: (1) You will need to use a scientific or graphing calculator to answer some of the questions. (2) You will have to decide whether to put your calculator in degree or radian mode for some problems. (3) All figures that accompany problems are plane figures unless otherwise stated. Figures are drawn as accurately as possible to provide useful information for solving the problem, except when it is stated in a particular problem that the figure is not drawn to scale. (4) Unless otherwise indicated, the domain of a function is the set of all real numbers for which the functional value is also a real number.
 Reference Information. The following formulas are provided for your information. Volume of a right circular cone with radius r and height h: Lateral area of a right circular cone if the base has circumference C and slant height is l: Volume of a sphere of radius r: Surface area of a sphere of radius r: Volume of a pyramid of base area B and height h:

1. x2/3 + x4/3 =

(A) x2/3

(B) x8/9

(C) x

(D) x2

(E) x2/3 (x2/3 + 1)

2. In three dimensions, what is the set of all points for which x = 0?

(A) the origin

(B) a line parallel to the x-axis

(C) the yz-plane

(D) a plane containing the x-axis

(E) the x-axis

3. Expressed with positive exponents only, is equivalent to

(A)

(B)

(C)

(D)

(E)

4. If f(x) = and g(x) = x3 + 8, find (f g)(3).

(A) 3.3

(B) 5

(C) 11

(D) 35

(E) 50.5

5. x > sin x for

(A) all x > 0

(B) all x < 0

(C) all x for which x 0

(D) all x

(E) all x for which – < x < 0

6. The sum of the zeros of f(x) = 3×2 – 5 is

(A) 3.3

(B) 1.8

(C) 1.7

(D) 1.3

(E) 0

7. The intersection of a plane with a right circular cylinder could be which of the following?

I. A circle

II. Parallel lines

III. Intersecting lines

(A) I only

(B) II only

(C) III only

(D) I and II only

(E) I, II, and III

8. There is a linear relationship between the number of cricket chirps and the temperature of the air. A biologist developed the regression model y = 24.9 + 3.5x, valid for values of x between 10 and 24. In this model, x is the number of chirps per minute and y is the predicted temperature in degrees Fahrenheit. What is the estimated increase in temperature that corresponds to an increase of 8 chirps per minute?

(A) 3.5°

(B) 24.9°

(C) 28°

(D) 28.4°

(E) 52.9°

9. The graph of f(x) = has a vertical asymptote at x =

(A) 0 only

(B) 5 only

(C) 10 only

(D) 0 and 5 only

(E) 0, 5, and 10

10. P(x) = x5 + x4 – 2×3 – x – 1 has at most n positive zeros. Then n =

(A) 0

(B) 1

(C) 2

(D) 3

(E) 5

11. Of the following lists of numbers, which has the largest standard deviation?

(A) 2, 7, 15

(B) 3, 7, 14

(C) 5, 7, 12

(D) 10, 11, 12

(E) 11, 11, 11

12. If f(x) is a linear function and f(2) = 1 and f(4) = –2, then f(x) =

(A)

(B)

(C)

(D)

(E)

13. The length of the radius of a circle is one-half the length of an arc of the circle. How large is the central angle that intercepts that arc?

(A) 60º

(B) 120º

(C) 1R

(D) 2R

(E) πR

14. If f(x) = 2x + 1, then f–1(7) =

(A) 2.4

(B) 2.6

(C) 2.8

(D) 3

(E) 3.6

15. Find all values of x that satisfy the determinant equation .

(A) –1

(B) –1 or 1.5

(C) 1.5

(D) –1.5

(E) –1.5 or 1

16. The 71st term of 30, 27, 24, 21, · · · , is

(A) 5325

(B) 240

(C) 180

(D) –180

(E) –183

17. If 0 < x < and tan 5x = 3, to the nearest tenth, what is the value of tan x?

(A) 0.5

(B) 0.4

(C) 0.3

(D) 0.2

(E) 0.1

18. If 4.05p= 5.25q, what is the value of ?

(A) –0.11

(B) 0.11

(C) 1.19

(D) 1.30

(E) 1.67

19. A cylinder has a base radius of 2 and a height of 9. To the nearest whole number, by how much does the lateral area exceed the sum of the areas of the two bases?

(A) 101

(B) 96

(C) 88

(D) 81

(E) 75

20. If cos 67° = tan x°, then x =

(A) 0.4

(B) 6.8

(C) 7.8

(D) 21

(E) 29.3

21. P(x) = x3 + 18x – 30 has a zero in the interval

(A) (0, 0.5)

(B) (0.5, 1)

(C) (1, 1.5)

(D) (1.5, 2)

(E) (2, 2.5)

22. The lengths of the sides of a triangle are 23, 32, and 37. To the nearest degree, what is the value of the largest angle?

(A) 71°

(B) 83°

(C) 122°

(D) 128°

(E) 142°

23. If f(x) = and g(x) = , find the domain of f g.

(A) x –1

(B) x 2

(C) x –1, x 2

(D) x –1, x 3

(E) x –1

24. Two cards are drawn from a regular deck of 52 cards. What is the probability that both will be 7s?

(A) 0.149

(B) 0.04

(C) 0.012

(D) 0.009

(E) 0.005

25. If =3.216, then =

(A) 321.6

(B) 32.16

(C) 10.17

(D) 5.67

(E) 4.23

26. What is the domain of the function ?

(A) –7.5 < x < 7.5

(B) x < –7.5 or x > 7.5

(C) x < –2.7 or x > 2.7

(D) x < –3.2 or x > 3.2

(E) x < 1.9 or x > 1.9

27. A magazine has 1,200,000 subscribers, of whom 400,000 are women and 800,000 are men. Twenty percent of the women and 60 percent of the men read the advertisements in the magazine. What is the probability that a randomly selected subscriber reads the advertisements?

(A) 0.30

(B) 0.36

(C) 0.40

(D) 0.47

(E) 0.52

28. Let S be the sum of the first n terms of the arithmetic sequence 3, 7, 11, · · · , and let T be the sum of the first n terms of the arithmetic sequence 8, 10, 12, · · · . For n > 1, S = T for

(A) no value of n

(B) one value of n

(C) two values of n

(D) three values of n

(E) four values of n

29. On the interval , the function has a maximum value of

(A) 0.78

(B) 1

(C) 1.1

(D) 1.2

(E) 1.4

30. A point has rectangular coordinates (3,4). The polar coordinates are (5,). What is the value of ?

(A) 30°

(B) 37°

(C) 51°

(D) 53°

(E) 60°

31. If f(x) = x2 – 4, for what real number values of x will f(f(x)) = 0?

(A) 2.4

(B) ±2.4

(C) 2 or 6

(D) ±1.4 or ±2.4

(E) no values

32. If f(x) = x log x and g(x) = 10x, then g(f(2)) =

(A) 24

(B) 17

(C) 4

(D) 2

(E) 0.6

33.If , then

(A) 1.4

(B) 1.5

(C) 1.6

(D) 2.0

(E) 2.7

34. The figure above shows the graph of 5x. What is the sum of the areas of the triangles?

(A) 32,550

(B) 16,225

(C) 2604

(D) 1302

(E) 651

35. (p,q) is called a lattice point if p and q are both integers. How many lattice points lie in the area between the two curves x2 + y2 = 9 and x2 + y2 – 6x + 5 = 0?

(A) 0

(B) 1

(C) 2

(D) 3

(E) 4

36. If sin A = , 90° < A < 180°, cos B = , and 270° < B < 360°, the value of sin (A + B) is

(A) –0.83

(B) –0.55

(C) –0.33

(D) 0.73

(E) 0.95

37. For all real numbers x, f(2x) = x2 – x + 3. An expression for f(x) in terms of x is

(A) 2×2 – 2x + 3

(B) 4×2 – 2x + 3

(C)

(D)

(E) x2 – x + 3

38. For what value(s) of k is x2 – kx + k divisible by x – k?

(A) only 0

(B) only 0 or

(C) only 1

(D) any value of k

(E) no value of k

39. If the graphs of x2 = 4(y + 9) and x + ky = 6 intersect on the x-axis, then k =

(A) 0

(B) 6

(C) –6

(D) no real number

(E) any real number

40. The length of the latus rectum of the hyperbola whose equation is x2 – 4y2 = 16 is

(A) 1

(B) 2

(C)

(D) 2

(E) 16

41. If

and f1 = 3, then f5 =

(A) 1

(B) 2

(C) 4

(D) 8

(E) 16

42. How many distinguishable rearrangements of the letters in the word CONTEST start with the two vowels?

(A) 120

(B) 60

(C) 10

(D) 5

(E) none of these

43. Which of the following translations of the graph of y = x2 would result in the graph of y = x2 – 6x + k, where k is a constant greater than 10?

(A) Left 6 units and up k units

(B) Left 3 units and up k + 9 units

(C) Right 3 units and up k + 9 units

(D) Left 3 units and up k – 9 units

(E) Right 3 units and up k – 9 units

44. How many positive integers are there in the solution set of ?

(A) 0

(B) 2

(C) 4

(D) 5

(E) an infinite number

45. During the year 1995 the price of ABC Company stock increased by 125%, and during the year 1996 the price of the stock increased by 80%. Over the period from January 1, 1995, through December 31, 1996, by what percentage did the price of ABC Company stock rise?

(A) 103%

(B) 205%

(C) 305%

(D) 405%

(E) 505%

46. If x0 = 3 and xn + 1, then x3 =

(A) 15.9

(B) 31.7

(C) 44.9

(D) 65.2

(E) 173.9

47. When the smaller root of the equation 3×2 + 4x – 1 = 0 is subtracted from the larger root, the result is

(A) –1.3

(B) 0.7

(C) 1.3

(D) 1.8

(E) 2.0

48. Each of a group of 50 students studies either French or Spanish but not both, and either math or physics but not both. If 16 students study French and math, 26 study Spanish, and 12 study physics, how many study both Spanish and physics?

(A) 4

(B) 5

(C) 6

(D) 8

(E) 10

49. If x, y, and z are positive, with xy = 24, xz = 48, and yz = 72, then x + y + z =

(A) 22

(B) 36

(C) 50

(D) 62

(E) 96

50. sin–1 (cos 100°) =

(A) –1.4

(B) –0.2

(C) 0.2

(D) 1.0

(E) 1.4

MODEL TEST 5

The following explanations are keyed to the review portions of this book. The number in brackets after each explanation indicates the appropriate section in the Review of Major Topics (Part 2). If a problem can be solved using algebraic techniques alone, [algebra] appears after the explanation, and no reference is given for that problem in the Self-Evaluation Chart at the end of the test.

An asterisk appears next to those solutions in which a graphing calculator is necessary.

1. (E) Factor out x2/3, the greatest common factor of x2/3 + x4/3, to get x2/3 + x4/3 = x2/3(1 + x2/3). [1.4]

2. (C) When x = 0, y and z can be any value. Therefore, any point in the yz-plane is a possible member of the set. [2.2]

3. (D) . [algebra]

4. * (A) Enter f in Y1 and g in Y2, return to the home screen and enter Y1(Y2(3)) to get the correct answer choice. [1.1]

5. * (A) Plot the graph of y = x – sin x and observe that the graph lies above the x-axis for all x > 0. [1.3]

6. * (E) Plot the graph of y = 3×2 – 5 in the standard window. The symmetry about the y-axis indicates that the zeros are opposites and therefore sum to zero.
An alternative solution is to use the fact that the zeros of a quadratic function sum to . [1.2]

7. (D) If the plane is perpendicular to the axis of the cylinder, the intersection will be a circle, so I is possible. If the plane is parallel to the axis of the cylinder, the intersection will be parallel lines, so II is possible. It is impossible for the intersection of a plane and a cylinder to form intersecting lines because there are no intersecting lines on a cylinder. [2.2]

8. (C) The value of the slope is the increase in temperature for each additional chirp. Therefore, multiply 8 chirps times 3.5° per chirp to get 28°. [4.1]

9. * (B) Plot f in a standard window and use TRACE to observe that y increases without bound on either side of x = 5.
An alternative solution can be found by observing that the denominator of f is (x – 5)2, which equals zero only when x = 5. [1.2]

10. * (B) Plot the function P in a standard window and observe that it crosses the x-axis only once to the right of the y-axis. Descartes’ Rule of Signs guarantees at most one zero because P(x) has only 1 sign change. [1.2]

11. (A) The standard deviation of a data set is a measure of the spread in the data. Of the five data sets presented, Choice A exhibits the greatest spread and therefore has the greatest standard deviation. [4.1]

12. (A) The slope of f(x) = . Using the point-slope form, f(x) – 1 = . Therefore, f(x) = .[1.2]

13. (D) s = r. 2r = r. = 2R. [1.3]

14. * (B) Plot the graphs of y = 2x + 1 in the standard window. Since f–1(7) is the value of x such that f(x) = 7, also plot the graph of y = 7. The correct answer choice is the point where these two graphs intersect.
An alternative solution is to solve the equation 2x + 1 = 7: x = log26 =. [1.4]

15. (B) Evaluate the 2 by 2 determinant to get 2×2 – x = 3. Then factor 2×2 – x – 3 into (2x – 3)(x + 1) and set each factor to zero to get x = or x = –1. [3.3]

16. * (D) Use LIST/seq to construct the sequence as seq(30 – 3x, x, 0,70) and store this sequence in a list (e.g., L1). On the Home Screen, enter L1(71) for the 71st term, –180.
An alternative solution is to use the formula for the nth term of an arithmetic sequence: t71 = 30 – (70)(3) = –180. [3.4]

17. * (C) Put your calculator in radian mode: 5x = tan–1 3 1.249. x 0.2498. Therefore, tan x = 0.255139 0.3. [1.3]

18. * (C) Take log4.05 of both sides of the equation, getting p = log4.055.25q, or p = qlog4.055.25. Dividing both sides by q and changing to base 10 yields . [1.4]

19. * (C) Area of one base = πr2 = 4π.
Lateral area = 2πrh = 36π.
Lateral area – two bases = 36π – 8π = 28π 87.96 88. [2.2]

20. * (D) Put your calculator in degree mode and evaluate tan–1(cos 67°). [1.3]

21. * (C) Plot the graph of y = x3 + 18x – 30 in a [0,3] by [–5,5] window, and use CALC/zero to locate a zero at x 1.48. [1.2]

22. * (B) The angle opposite the 37 side (call it A) is the largest angle. By the law of cosines, 372 = 232 + 322 – 2(23)(32) cos A.

cos .
Therefore, A = cos–1 (0.125) 83°. [1.3]

23. (D) The domain of g is x –1 because x + 1 0. The domain of f g consists of all x such that g(x) is in the domain of f. Since g(3) = 2 and 2 is not in the domain of f, 3 must be excluded from the domain of f g. [1.1]

24. * (E) There are four 7s in a deck, and so P(first draw is a 7) = . There are now only three 7s among the remaining 51 cards, and so P(second draw is a 7) = . Therefore, P(both draws are 7s) = . [4.2]

25. * (C) Since = 3.216, y 10.34, and 10y 103.4, so 10.17. [algebra]

26. * (C) Plot the graph of in the standard window and zoom in once. Use the TRACE to determine that there are no y values on the graph between the approximate x values of –2.7 and 2.7.
An alternative solution is to use the fact that the domain of the square root function is x 0 and its range is y 0. However, since the domain of the log function is x > 0, the domain of the function f must satisfy 2×2 – 15 > 0, or x2 > 7.5. The approximate solution to this inequality is the correct answer choice C. [1.4]

27. * (D) Twenty percent of 400,000 women is 80,000, and 60% of 800,000 men is 480,000. Altogether 560,000 subscribers read ads, or about 47%. [4.1]

28. (B) Equate S and T using the formula for the sum of the first n terms of an arithmetic series: . This equation reduces to n2 – 6n = 0, for which 6 is the only positive solution. [3.4]

29. * (D) Plot the graph of in a by [–2,2] window, and observe that the maximum value of y occurs at the endpoints. Return to the home screen, and enter Y1(π/4) to get the correct answer choice.
An alternative solution uses the fact that on the interval, the maximum value of Therefore, the maximum value of f(x) is. [1.3]

30. * (D) sin
Therefore, [1.3, 2.1]

31. * (D) Plot the graph of f(f(x)) in the standard window by entering Y1 = x2 – 4, Y2 = Y1(x) and de-selecting Y1. The graph has 4 zeros located symmetrically about the y-axis, making answer choice D the only possible one.
An alternative solution is to evaluate f(f(x)) = (x2 – 4)2 – 4 = 0 and solve by setting, or x2 = 6 or 2, so or . Again, D is the only answer choice with 4 solutions. [1.2]

32. * (C) Enter f into Y1 and g into Y2. Evaluate Y2(Y1(2)) for the correct answer choice.
An alternative solution is to use the properties of logs to evaluate f(2) = 2 log 2 = log 4 and g(log 4) = 10log 4 = 4 without a calculator. [1.4]

33. * (B) . [1.4]

34. * (D) The width of each rectangle is 2 and the heights are 50 = 1, 52 = 25, and 54 = 625. Therefore, the total area is 2(1 + 25 + 625) = 1302. [1.4]

35. * (D) Plot the graphs of and in the standard window, but with FORMAT set to GridOn. The “grid” consists exactly of the lattice points. ZOOM/ZBox around the area enclosed by the two graphs, and count the number of lattice points in that area to be 3. The points (1,0) and (3,0) appear close to the boundary, but a mental check finds that (1,0) is on the boundary of the second curve, while (3,0) is on the boundary of the first. [2.1]

36. * (E) First observe the facts that A is in Quadrant II and B is in Quadrant IV imply that A + B is in Quadrant I or II, so that sin (A + B) > 0. With your calculator either in degree or radian mode, enter sin(sin–1 (3/5) + cos–1 (1/3)) to get the correct answer choice.
An alternative solution is to use the formula for the sin of a sum of two angles. Using sin2x + cos2x = 1 and the fact that A is in Quadrant I, together with sin A =, implies cos A =. Similarly, cos B = and B in Quadrant IV implies sin B = .
Substituting these values into sin (A + B) = sinAcosB + cosAsinB yields the correct answer choice. [1.3]

37. (C) . [1.1]

38. (A) Using the factor theorem, substitute k for x and set the result equal to zero. Then k2 – k2 + k = 0, and k = 0. [1.2]

39. (E) If the graphs intersect on the x-axis, the value of y must be zero. Since the value of y is zero, it does not matter what k is. [1.2]

40. (B) a2 = 16. b2 = 4. Latus rectum = = 2. [2.1]

41. (D) [3.4]

42. * (A) There are 5 consonants, CNTST, but the two Ts are indistinguishable, so there are = 60 ways of arranging these. There are two ways of arranging the 2 vowels in the front. Therefore, there are 2 · 60 = 120 distinguishable arrangements. [3.1]

43. (E) Complete the square on x2 – 6x by adding 9. Then x2 – 6x + k = x2 – 6x + 9 + k – 9 = (x – 3)2 + (k – 9). This expression represents the translation of x2 by 3 units right and k – 9 units up. [2.1]

44. * (A) Plot the graph of in the standard window and observe the vertical asymptote at where this graph lies above the x-axis, and there are no integer values of x in this interval.
An alternative solution is to solve the equation, or .
If you test values in the intervals, , and you find that only the middle interval satisfies the inequality, but it contains no integers. [1.2]

45. * (C) Let the starting price of the stock be \$100. During the first year a 125% increase means a \$125 increase to \$225. During the second year an 80% increase of the \$225 stock price means a \$180 increase to \$405. Thus, over the 2-year period the price increased \$305 from the original \$100 starting price. Therefore, the price increased 305%. [algebra]

46. * (D) Enter 3 into your calculator. Then enter Ans three times to accomplish three iterations that result in x3 and the correct answer choice. [3.4]

47. * (D) Use your calculator program for the Quadratic Formula to find the roots of the equation. Then subtract the smaller root (–1.54) from the larger (0.22) and round to get the correct answer choice.
An alternative solution is to substitute the values of a, b, and c into the Quadratic Formula to find the algebraic solutions; then subtract the smaller from the larger to find the decimal approximation to the answer:. [1.2]

48. (A)

a + b + c + 16 = 50, a + b = 26, a + c = 12. Subtracting the first two equations and then the first and third gives c = 8, b = 22, and a = 4. Four students take both Spanish and physics. [3.1]

49. (A) First evaluate the ratio . Cross-multiply to get z = 2y, and substitute in the third equation: yz = y(2y) = 2y2 = 72. Therefore, y = 6, z = 12, and x = 4, and x + y + z = 22. [1.2]

50. * (B) With your calculator in degree mode, evaluate sin–1(cos(100)) = –10° directly. To change to radians, return your calculator to radian mode and key in –10° (using ANGLE/°). This will return –0.17.
An alternative solution is to convert –10° to radians by multiplying by. [1.3]

Self-Evaluation Chart for Model Test 5

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